The volume of an open top box is created by cutting a square from each corner of a 10 in. by 12 in. piece of paper. The situation is represented by the function V(x)=x(10−2x)(12−2x)

Question

. Find the maximum volume of a box that can be created from the piece of paper. Round the answer to one decimal place.

(1 point)
cubic inches

Answer 1

To find the maximum volume, we need to find the maximum value of the function V(x) = x(10-2x)(12-2x).

Taking the first derivative of V(x) with respect to x, we get:

V'(x) = (10-2x)(12-2x) + x(-2)(12-2x) + x(10-2x)(-2)

Simplifying, we get:

V'(x) = (10-2x)(12-2x) - 2x(12-2x) - 2(10-2x)(12-2x)

Expanding further, we have:

V'(x) = 120 - 24x - 20x + 4x^2 - 24x + 4x^2 - 60 + 12x + 60 - 4x^2

Combining like terms, we get:

V'(x) = -28x + 120

To find the critical points of the function, we set V'(x) equal to zero and solve for x:

-28x + 120 = 0
-28x = -120
x = 120/28
x ≈ 4.3

We also know that the maximum volume will occur either at the critical point or at the endpoints of the interval [0, 5] (since we can't have negative dimensions for the box).

To check the endpoints, we evaluate V(x) at x = 0 and x = 5:

V(0) = 0(10-2*0)(12-2*0) = 0
V(5) = 5(10-2*5)(12-2*5) = 5(10-10)(12-10) = 0

Both endpoints yield a volume of 0, so the maximum volume must occur at the critical point x ≈ 4.3.

To find the maximum volume, we substitute x ≈ 4.3 back into the original function:

V(x) = x(10-2x)(12-2x)
V(4.3) ≈ 4.3(10-2*4.3)(12-2*4.3) ≈ 4.3(10-8.6)(12-8.6) ≈ 4.3(1.4)(3.4) ≈ 20.3 cubic inches

Therefore, the maximum volume of the box that can be created from the piece of paper is approximately 20.3 cubic inches, rounded to one decimal place.